circular_counter.py

"""
There are people sitting in a circular fashion,
print every third member while removing them,
the next counter starts immediately after the member is removed.
Print till all the members are exhausted.

For example:
Input: consider 123456789 members sitting in a circular fashion,
Output: 369485271
"""

a = ['1','2','3','4','5','6','7','8','9']

def josepheus(int_list, skip):
  skip = skip - 1 #list starts with 0 index
  idx = 0
  len_list = (len(int_list))
  while len_list>0:
    idx = (skip+idx)%len_list #hashing to keep changing the index to every 3rd
    print(int_list.pop(idx))
    len_list -= 1


josepheus(a,3)

"""
the reason for hashing is that we have to find the index of the item which needs to be removed.
So for e.g. if you iterate with the initial list of folks with every 3rd item eliminated:

INPUT
int_list = 123456789
skip = 3

While Iteration:

int_list = 123456789
len(int_list) = 9
skip = 2 # as int_list starts from 0
idx = (0 + 2) % 9 #here previous index was 0
so 3rd element which is 3 in this case eliminated
int_list = 12456789
len(int_list) = 8
idx = (2 + 2) % 8 #here previous index was 2
so 3rd element starting from 4th person which is 6 would be deleted.
and so on
The reason why we have to do this way is I am not putting the people who escape at the back of list so ideally in 2 while iteration the list should have been
45678912 and then no hashing needed to be done, which means you can directly remove the third element
"""